That is the problem I try to solve few days now but not achieve anything. Anyone has any ideas, please share. Thank you.

P.S: It is not an assignment, and sorry for my bad English.

Supplement 1: I think I need to make it clear: The problem I suggested is about to find $f$ which satisfies $(5)$. I also show the way I find the problem, and the only purpose of which is that it may provide ideas for solving.

Supplement 2: I think I have proved the identity of $f(n,3)$ in the comment below
$$f(n,3) = \sum_{i=0}^n \binom{n}{i}^2 \binom{2i}{i} \tag 6$$
by using Double Counting:

We double-count the number of ways to choose a sex-equal subgroup, half of which are on the special duty, from the group which includes $n$ men and $n$ women (the empty subgroup is counted).

The first way of counting: The number of satisfying subgroups which contain $2i$ people is $\binom{n}{i}^2 \binom{2i}{i}$. So we have the number of satisfying subgroups is $RHS(6)$.

The second way of counting: The number of satisfying subgroups which contain $2(a_2+a_3)$ people, $a_2$ women on the duty and $a_3$ men on the duty is
$$\left ( \frac{n!}{a_1!a_2!a_3!} \right )^2$$.
So the number of satisfying subgroups is $LHS(6)$.

Did you try to calculate some values and ask OEIS?
–
draks ...Apr 23 '12 at 14:27

3

The square in the identity is a red herring; it becomes much easier to understand when you write it as $\sum_i {n \choose i} {n \choose n-i} = {2n \choose n}$, and then the appropriate generalization is Vandermonde's identity: en.wikipedia.org/wiki/Vandermonde's_identity
–
Qiaochu YuanApr 23 '12 at 15:15

@draks: I have just tried with the case of $k=3$ and had this oeis.org/A002893: $f(n,3) = \sum_{i=0}^n \binom{n}{i}^2 \binom{2i}{i}$.
–
Vincent J. RuanApr 23 '12 at 15:57

1

@Vincent: What's "it"? The identity or the link? And what's "here"? Your browser or this question? By the way, Vandermonde's identity further generalizes to the Rothe-Hagen identity. Also by the way, you can get displayed equations by using double dollar signs instead of single dollar signs. They look nicer and are easier to read.
–
jorikiApr 23 '12 at 16:21

1

@Vincent: You can get equation tags right-aligned using e.g. \tag 6. That gets them out of the way of the equations themselves.
–
jorikiApr 24 '12 at 11:58

I think it's not about right or wrong when considering a generalization. In fact, both the generalization Qiaochu Yuan and you suggested (call it "Vandermonde generalization" from now on) and mine are just two different ways to generalize $(4)$. As you said, the Vandermonde generalization has its basis, and so is mine.
–
Vincent J. RuanApr 24 '12 at 6:41